Why does Swift return an unexpected pointer when converting an optional String into an UnsafePointer?

Question

I noticed some unusual behaviour when working with a C library which took strings in as const char * (which is converted to Swift as UnsafePointer

Answer 1

As mentioned in the comments, this is a clear bug in Swift.

Here's a workaround I'm using. If you can't trust Swift to convert the strings to pointers for you, then you've got to do it yourself.

Assuming C function defined as:

void multiString(const char *arg0, const char *arg1, const char *arg2);

Swift code:

func callCFunction(arg0: String?, arg1: String?, arg2: String?) {
    let dArg0 = arg0?.data(using: .utf8) as NSData?
    let pArg0 = dArg0?.bytes.assumingMemoryBound(to: Int8.self)

    let dArg1 = arg1?.data(using: .utf8) as NSData?
    let pArg1 = dArg1?.bytes.assumingMemoryBound(to: Int8.self)

    let dArg2 = arg2?.data(using: .utf8) as NSData?
    let pArg2 = dArg2?.bytes.assumingMemoryBound(to: Int8.self)

    multiString(pArg1, pArg2, pArg3)
}

Warning:

Don't be tempted to put this in a function like:

/* DO NOT USE -- BAD CODE */
func ocstr(_ str: String?) -> UnsafePointer? {
    guard let str = str else {
        return nil
    }

    let nsd = str.data(using: .utf8)! as NSData

    //This pointer is invalid on return:
    return nsd.bytes.assumingMemoryBound(to: Int8.self)
}

which would remove repeated code. This doesn't work because the data object nsd gets deallocated at the end of the function. The pointer is therefore not valid on return.